Question

an object is thrown upward with a velocity of 40m/s and has acceleration 10m/s^2 ..find the height obtained by the object​

Answer 1

Explanation:

Initial Velocity u=40

Fianl velocity v=0

Height, s=?

By third equation of motion

v

2

−u

2

=2gs

0−40

2

=−2×10×s

s=

20

160

⇒s=80m/s

Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m

Total Diaplacement =0, Since the initial and final point is same.

Answer 2

According to the equation of the motion under gravity
v
2
−u
2
=2gs
u=initial velocity of the stone=40m/s
v= Final velocity of the stone=0m/s
Let h be the maximum height attained by the stone
Therefore,
0
2
−40
2
=2(−10)h
h=(40×40)/20=80
Therefore, total distance covered by the stone during its upward and downward journey=80+80=160m
Net displacement during its upward and downward journey=80+(−80)=0